A) Find the time to reach the maximum height. The x-component and y-component of the motion during flight can be treated as independent of each other. At maximum height, the y-component of velocity is momentarily zero. Use the expression for motion in the y-direction: vy = vy0 − gt and set vy to zero at the time ty max of maximum height: 0 = v0 sin θ − gty max. Then solve for ty max: ty max = v0 sin θ g. Suppose the initial speed is 50 m/s and the launch angle is θ = 53°. In this case we find ty max = s. (B) Find the maximum height ymax. The distance the particle moves upward while slowing with the acceleration −g is ymax = vy0ty max − 1 2g(ty max)2. Therefore, substitute in the expression for ty max from part (a), and simplify, to obtain ymax = v02 sin2 θ 2g. Suppose the initial speed is 50 m/s and the launch angle is θ = 53°. In this case, we have ymax = m. (C) Find the entire time of flight. There are at least two approaches. One approach is to use the y-displacement equation, noting that the projectile starts from the ground and ends on the ground, where y is zero. y = vy0t − 1 2gt2 hence, when the projectile has landed, 0 = v0 sin θ (tflight) − 1 2g(tflight)2. Cancel a factor of t from both terms, rearrange, and solve: tflight = 2v0 sin θ g. An easier way is to recognize the symmetry in this motion: the total time of flight is twice the time to reach the highest point. Check to see if this is so! Suppose the initial speed is 50 m/s and the launch angle is θ = 53°. In this case, we have: tflight = s.
(A) Find the time to reach the maximum height.
The x-component and y-component of the motion during flight can be treated as independent of each other. At maximum height, the y-component of velocity is momentarily zero. Use the expression for motion in the y-direction:
and set vy to zero at the time ty max of maximum height:
Then solve for ty max:
| v0 sin θ |
| g |
Suppose the initial speed is 50 m/s and the launch angle is θ = 53°. In this case we find
(B) Find the maximum height ymax.
The distance the particle moves upward while slowing with the acceleration −g is
| 1 |
| 2 |
Therefore, substitute in the expression for ty max from part (a), and simplify, to obtain
| v02 sin2 θ |
| 2g |
Suppose the initial speed is 50 m/s and the launch angle is θ = 53°. In this case, we have
(C) Find the entire time of flight.
There are at least two approaches. One approach is to use the y-displacement equation, noting that the projectile starts from the ground and ends on the ground, where y is zero.
| 1 |
| 2 |
hence, when the projectile has landed,
| 1 |
| 2 |
Cancel a factor of t from both terms, rearrange, and solve:
| 2v0 sin θ |
| g |
An easier way is to recognize the symmetry in this motion: the total time of flight is twice the time to reach the highest point. Check to see if this is so!
Suppose the initial speed is 50 m/s and the launch angle is θ = 53°. In this case, we have:
(D) Find the range.
While in flight, the projectile's horizontal velocity component is constant and x attains the maximum value when t = tflight
Call this maximum value the range R
Substituting the expression for tflight found above and rearranging then gives
| 2v02sin θ cos θ |
| g |
Suppose the initial speed is 50 m/s and the launch angle is θ = 53°. Find R.
(E) Find the angle that gives the maximum range.
To find the maximum range, we will use the trigonometric identity:
Substituting this into the above expression for R gives
| v02 sin(2θ) |
| g |
The maximum R occurs when sin(2θ) is at its maximum, which occurs at
Use the Active Figure to verify this.
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